It's definitely possible.
Considering it's a Full Binary Tree, once the number of nodes is determined, theoretically, the shape of tree is unique.
Deem the level order traversal as an array, for example, 1 2 3 4 5 6 7.
It represents such tree:
1
2 3
4 5 6 7
What you want to get is the post order traversal: 4 5 2 6 7 3 1.
The first step is calculate how deep the tree was:
depth = ceil(log(2, LevelOrderArray.length)) // =3 for this example
after that, set up a counter = 0, and extract nodes from the bottom level of the original array, one by one:
for(i=0, i<LevelOrderArray.length, i++){
postOrderArray[i] = LevelOrderArray[ 2 ^ (depth-1) +i ] //4,5,....
counter++; //1,2,.....
}
But notice that once the counter can be divided by 2, that means you need to retrieve another node from upper level:
if(counter mod 2^1 == 0)
postOrderArray[i] = LevelOrderArray[ 2 ^ (depth -2) + (++i) ] // =2 here,
//which is the node you need after 4 and 5, and get 3 after 6 and 7 at the 2nd round
Don't ++ the counter here, because the counter represents how many nodes you retrieved from the bottom level.
Every time 2^2 = 4 nodes was pop out, retrieve another node from 3rd level (counting from bottom)
if(counter mod 2^2 == 0)
postOrderArray[i] = LevelOrderArray[ 2 ^ (depth -3) + (++i) ] // =1
Every time 2^3 = 8 nodes was pop out, again, retrieve another node from 4th level
.... until the loop is finished.
It's not strict C++ code, only the concept. If you fully understand the algorithm, the value of tree nodes doesn't matter at all, even though there are all 0 and 1. The point is although you didn't build up the tree in program, but build it up in your mind instead, and convert it into algorithm.